# Natural Transformations

“I didn’t invent categories to study functors; I invented them to study natural transformations.” – Saunders Mac Lane, one of the founders of category theory

A natural transformation is, loosely speaking, a homomorphism between functors. Its definition may seem strange but the example after it should illustrate its use.

Definition

Let $F, G : \mathcal C\to \mathcal D$ be two covariant functors. A natural transformation

$T:F\Rightarrow G$

assigns, to each object $A\in \mathcal C$, a morphism $T_A : F(A) \to G(A)$ such that for all morphisms $f:A\to B$ in $\mathcal C$, we have

$G(f) \circ T_A = T_B \circ F(f) : F(A) \longrightarrow G(B).$

In diagram, we have the following, where the right square commutes.

In summary a natural transformation from one functor to another takes an object in the base category to a morphism between the target objects.

### Example

Let $GL_2(-)$ and $U(-)$ be functors $A\text{-}\mathbf{Alg} \to \mathbf{Gp}$ given by

• $GL_2 B$ is the group of invertible 2 × 2 matrices with entries in B,
• $U(B)$ is the group of units in B.

Then there is a natural transformation $\det : GL_2(-) \Rightarrow U(-)$: for each A-algebra B we take:

$\det_B : GL_2(B) \to U(B), \quad \begin{pmatrix} a & b \\ c & d \end{pmatrix} \mapsto ad - bc.$

Naturality means if $f:B\to C$ is a homomorphism of A-algebras, then the following commutes:

which is easily verified.

### Exercise A

Suppose we have a morphism $f : A\to A'$ in the category $\mathcal C$. From f define a natural transformation between the functors

$T(f) : \mathrm{hom}_{\mathcal C}(A', -) \Rightarrow\mathrm{hom}_{\mathcal C}(A, -).$

# Composing Natural Transformations

Definition.

We can compose natural transformations as follows. For covariant functors:

$F, G, H : \mathcal C \to \mathcal D$,

suppose we have natural transformations $T : F\Rightarrow G$ and $U:G\Rightarrow H$. We can compose $U\circ T : F\Rightarrow H$ by

$(A \in \mathcal C) \mapsto U_A\circ T_A : F(A) \to H(A).$

If we can find natural transformations $T : F\Rightarrow G$ and $U:G\Rightarrow F$ such that

$A \in \mathcal C \implies U_A \circ T_A = 1_{F(A)}, \ T_A \circ U_A = 1_{G(A)}$,

then T and U are called natural isomorphisms.

Functors $F, G: \mathcal C \to \mathcal D$ are said to be naturally isomorphic if there exist natural isomorphisms between them as above.

### Example 1

Take functors $GL_2(-), U(-): A\text{-}\mathbf{Alg} \to \mathbf{Gp}$ as above. We defined the natural transformation

$\det : GL_2(-) \Rightarrow U(-), \quad B \mapsto \det : GL_2 B \to U(B),$

which takes the determinant of the matrix. For each integer n we also have the following natural transformation

$p_n : U(-) \Rightarrow U(-), \quad B \mapsto (p_n : U(B) \to U(B), x \mapsto x^n).$

Composing $p_n \circ \det$ then takes each A-algebra B to the map

$GL_2 B \to U(B), \quad \begin{pmatrix} a & b \\ c & d\end{pmatrix} \mapsto (ad - bc)^n.$

Even more simply, we have $p_m \circ p_n = p_{mn}$.

### Example 2

Let $\mathcal C = \mathbf{FinVec}_k$ be the category of finite-dimensional vector spaces over a fixed field k; the morphisms are the linear maps. Consider the functors:

$1_{\mathcal C}, F : \mathcal C \longrightarrow \mathcal C$,

where $1_{\mathcal C}$ is the identity functor and F takes the vector space V to its double-dual $V^{\vee\vee}$ (recall that $V^{\vee}$ is the space of all linear maps $V\to k$). Each linear map $f:V\to W$ gives $f^\vee : W^\vee \to V^\vee$ and thus $f^{\vee\vee} : V^{\vee\vee} \to W^{\vee\vee}$.

It follows from linear algebra that we have a natural isomorphism

$\varphi_V : V \stackrel \cong \longrightarrow V^{\vee\vee}, \quad v \mapsto (\alpha \in V^{\vee} \mapsto \alpha(v)).$

One can visualize this as taking inner product $\left< \alpha, v\right>$ for $\alpha \in V^\vee$ and $v\in V$. Fixing $\alpha$ gives a linear $V\to k$ while fixing $v$ gives a linear $V^\vee \to k$, i.e. an element of $V^{\vee\vee}$.

Hence we get a natural isomorphism $T : 1_{\mathcal C} \Rightarrow F$ which takes $V$ to $\varphi_V : V \to V^{\vee\vee}$.

### Optional Exercise

Define another composition of natural transformations as follows. Suppose $F, G : \mathcal C \to \mathcal D$ and $F', G' : \mathcal D \to \mathcal E$ are covariant functors. Suppose also $T : F\Rightarrow G$ and $U : F' \Rightarrow G'$ are natural transformations. Define a natural transformation

$U * T : F' \circ F \Rightarrow G' \circ G,\quad F' \circ F,\ G' \circ G : \mathcal C \to \mathcal E.$

Hint: in diagram form we have:

# Examples

Earlier, we defined an isomorphism

$\overbrace{\mathrm{Hom}_{A\text{-alg}}(A[X, Y]/(XY - 1), B)}^F \cong U(B),$

where U(B) returns the set of units of B. We can now prove this is natural using category theory. Note that both sides are functors $A\text{-}\mathbf{Alg} \to \mathbf{Set}$ in terms of B. We will define a natural transformation $T : F \Rightarrow U(-)$:

$A\text{-algebra } B \implies \begin{cases}T_B : \mathrm{Hom}_{A\text{-alg}}(A[X, Y]/(XY - 1), B) \to U(B)\\ (f : A[X, Y]/(XY - 1) \to B) \mapsto f(X).\end{cases}$

To prove that this is a natural isomorphism, we can define a reverse natural transformation $T' : U(-) \Rightarrow F$. However, there is a simpler way. We note that for each A-algebra B, the map $T_B$ is bijective.

• Indeed, any invertible element $\alpha \in U(B)$ gives us an A-algebra homomorphism $f:A[X,Y]/(XY - 1) \to B$ by taking $X, Y$ to $\alpha, \alpha^{-1}$ respectively.

Now apply the following to complete the proof.

Proposition 1.

Suppose $F, G:\mathcal C\to \mathcal D$ are functors and $T: F\Rightarrow G$ is a natural transformation. If, for each $A\in \mathcal C$, the morphism $T_A$ in $\mathcal D$ is an isomorphism, then $T$ is a natural isomorphism.

Proof

We need to find an inverse natural transformation $T' : G\Rightarrow F$. For each $A\in \mathcal C$, let $T'_A : G(A) \to F(A)$ be the inverse of $T_A : F(A) \to G(A)$. To check that T’ is natural:

• Indeed, for any $f:A\to B$ in $\mathcal C$, we have $G(f) \circ T_A = T_B \circ F(f)$.
• Since $T'_A = T_A^{-1}$ we have $T_B' \circ G(f) = F(f) \circ T_A'$, which is exactly what we want. ♦

# Representable Functors

Thus, we have shown that the functor $U : A\text{-}\mathbf{Alg} \to \mathbf {Set}$ which takes C to its set of units can be represented by a hom functor $\mathrm{Hom}_{A\text{-alg}}(B, -)$ for some A-algebra B.

Definition.

Let $\mathcal C$ be a category and $F:\mathcal C\to \mathbf{Set}$ be a (covariant) functor. We say F is representable if there is an object $A \in \mathcal C$ such that

$F \cong \mathrm{hom}_{\mathcal C}(A, -)$ as a natural isomorphism.

For the case of a contravariant $F$, we require $A$ to satisfy $F \cong \mathrm{hom}_{\mathcal C}(-, A)$.

Exercise B

Prove that for a representable F, the object A is unique up to isomorphism.

It turns out a large class of functors are representable.

1. Consider the forgetful functor $F : \mathbf{Gp} \to \mathbf{Set}$. This is representable, for we can take the the group $\mathbb Z \in \mathbf{Gp}$. The natural transformation $T_G : \mathrm{hom}_{\mathbf{Gp}}(\mathbb Z, G) \to G$ then takes a group homomorphism $f:\mathbb Z \to G$ to the image $f(1)$. Clearly $T_G$ is a bijection of sets.

As an exercise, show that for each of the following categories, the forgetful functor which takes an object to its underlying set (and morphism to its underlying function), is representable.

$\mathbf{Ring}, \ A\text{-}\mathbf{Mod}, \ A\text{-}\mathbf{Alg},\ \mathbf{Top}.$

What about the category of non-commutative rings?

2. Consider $F:\mathbf{Gp} \to \mathbf{Set}$,

$F(G) = \{x \in G : x^2 = e\}.$

Note that this is not a subgroup in general. If $f:G\to H$ is a group homomorphism and $x\in F(G)$, then $f(x)^2 = f(x^2) = e$ so $f(x)\in F(H)$ so F is indeed a functor. F is representable since

$F \cong \mathrm{hom}_{\mathbf {Gp}}(\mathbb Z/(2), -)$.

3. Consider $F : A\text{-}\mathbf{Alg} \to \mathbf{Set}$ which takes an A-algebra B and returns the set $\{ (x,y) \in B \times B : x^2 + y^2 = 1\}$. F is representable since

$F \cong \mathrm{hom}_{A\text{-}\mathbf{Alg}}(A[X, Y]/(X^2 + Y^2 - 1), -)$.

### Optional Note

Let $I = [0, 1]$ be the closed interval in $\mathbf{Top}$. The functor $\mathrm{hom}_{\mathbf{Top}}(I, -)$ takes a space X to the set of all paths in X. We call this the path space of X. Similarly, if $S^1$ is the unit circle, $\mathrm{hom}_{\mathbf{Top}}(S^1, -)$ takes X to the set of all loops in X; this is called the loop space of X. To induce a canonical topology on these spaces, topologists prefer to work in a rather unusual category of topological spaces, those which are compactly generated and weakly Hausdorff, which allows them to define a very nice topology for function spaces. But that’s another story for another day.

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